Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{-2x + 10}{x^2 - 4x - 5} \div \dfrac{x + 6}{7x + 7} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-2x + 10}{x^2 - 4x - 5} \times \dfrac{7x + 7}{x + 6} $ First factor the quadratic. $q = \dfrac{-2x + 10}{(x + 1)(x - 5)} \times \dfrac{7x + 7}{x + 6} $ Then factor out any other terms. $q = \dfrac{-2(x - 5)}{(x + 1)(x - 5)} \times \dfrac{7(x + 1)}{x + 6} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -2(x - 5) \times 7(x + 1) } { (x + 1)(x - 5) \times (x + 6) } $ $q = \dfrac{ -14(x - 5)(x + 1)}{ (x + 1)(x - 5)(x + 6)} $ Notice that $(x - 5)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -14(x - 5)\cancel{(x + 1)}}{ \cancel{(x + 1)}(x - 5)(x + 6)} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $q = \dfrac{ -14\cancel{(x - 5)}\cancel{(x + 1)}}{ \cancel{(x + 1)}\cancel{(x - 5)}(x + 6)} $ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ $q = \dfrac{-14}{x + 6} ; \space x \neq -1 ; \space x \neq 5 $